Difference between revisions of "2020 AIME I Problems/Problem 5"
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== Solution 2 (Inspired by 2018 CMIMC combo round) == | == Solution 2 (Inspired by 2018 CMIMC combo round) == | ||
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== Video Solution == | == Video Solution == | ||
https://www.youtube.com/watch?v=5iwdFd2OLKM&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=5 ~ MathEx | https://www.youtube.com/watch?v=5iwdFd2OLKM&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=5 ~ MathEx | ||
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+ | https://www.youtube.com/watch?v=E6YJh7vsLPU | ||
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+ | https://youtu.be/ctVf7X6fTLI | ||
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==See Also== | ==See Also== |
Revision as of 01:28, 16 February 2021
Contents
[hide]Problem
Six cards numbered through
are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.
Solution 1
Realize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two.
If we choose any of the numbers through
, there are five other spots to put them, so we get
. However, we overcount some cases. Take the example of
. We overcount this case because we can remove the
or the
. Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract
cases (namely,
,) to get
, but we have to add back one more for the original case,
. Therefore, there are
cases. Multiplying by
gives the desired answer,
.
-molocyxu
Solution 2 (Inspired by 2018 CMIMC combo round)
Similar to above, a correspondence between ascending and descending is established by subtracting each number from
.
We note that the given condition is equivalent to "cycling" for a contiguous subset of it. For example,
It's not hard to see that no overcount is possible, and that the cycle is either "right" or
"left." Therefore, we consider how many elements we flip by. If we flip
or
such elements, then there is one way to cycle them. Otherwise, we have
ways. Therefore, the total number of ascending is
, and multiplying by two gives
~awang11
Solution 3
Similarly to above, we find the number of ascending arrangements and multiply by 2.
We can choose cards to be the ascending cards, therefore leaving
places to place the remaining card. There are
to do this. However, since the problem is asking for the number of arrangements, we overcount cases such as
. Notice that the only arrangements that overcount are
(case 1) or if two adjacent numbers of
are switched (case 2).
This arrangement is counted
times. Each time it is counted for any of the
numbers selected. Therefore we need to subtract
cases of overcounting.
Each time
adjacent numbers of switched, there is one overcount. For example, if we have
, both
or
could be removed. Since there are
possible switches, we need to subtract
cases of overcounting.
Therefore, we have total arrangements of ascending numbers. We multiply by two (for descending) to get the answer of
-PCChess
Solution 4 (No overcounting)
Like in previous solutions, we will count the number of ascending arrangements and multiply by 2.
First, consider the arrangement 1-2-3-4-5-6. That gives us 1 arrangement which works.
Next, we can switch two adjacent cards. There are 5 ways to pick two adjacent cards, so this gives us 5 arrangements.
Now, we can "cycle" 3 adjacent cards. For example, 1-2-3 becomes 2-3-1 which becomes 3-1-2. There are 4 ways to pick a set of 3 adjacent cards, so this gives us 4x2=8 arrangements.
Cycling 4 adjacent cards, we get the new arrangements 2-3-4-1 (which works,) 3-4-1-2 (which doesn't work,) and 4-1-2-3 (which does work.) We get 6 arrangements.
Similarly, when cycling 5 cards, we find 2x2=4 arrangements, and when cycling 6 cards, we find 2x1=2 arrangements.
Adding, we figure out that there are 1+5+8+6+4+2=26 ascending arrangements. Multiplying by 2, we get the answer -i8Pie
Solution 5 (Official MAA 1)
First count the number of permutations of the cards such that if one card is removed, the remaining cards will be in ascending order. There is such permutation where all the cards appear in order:
There are
such permutations where two adjacent cards are interchanged, as in
The other such permutations arise from removing one card from
and placing it in a position at least two away from its starting location. There are
such positions to place each of the cards numbered
and
and
such positions for each of the cards numbered
and
This accounts for
permutations. Thus there are
permutations where one card can be removed so that the remaining cards are in ascending order. There is an equal number of permutations that result in the cards' being in descending order. This gives the total
.
Solution 6 (Official MAA 2)
More generally, suppose there are cards numbered
arranged in ascending order. If any one of the
cards is removed and placed in one of the
positions in the arrangement, the resulting permutation will have the property that one card can be removed so that the remaining cards are in ascending order. This accounts for
permutations. However, the original ascending order has been counted
times, and each order that arises by switching two neighboring cards has been counted twice. Hence the number of arrangements where one card can be removed resulting in the remaining cards' being in ascending order is
When
, this is
, and the final answer is
.
Solution 7 (Casework)
For ascending, if the goes in anything but the first two slots, the rest of the numbers have to go in ascending from
, which are
cases if there are
cards. If
goes in the second spot, then you can put any of the rest in the first slot but then the rest are determined, so in the case of
cards, that gives
more. If
goes in the first slot, that means that you are doing the same problem with
cards. So the recursion is
. There's
and
, so you get
,
,
, and
. Or you can see that
. We double to account for descending and get
. ~ahclark11
Video Solution
https://www.youtube.com/watch?v=5iwdFd2OLKM&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=5 ~ MathEx
Video Solution
https://www.youtube.com/watch?v=E6YJh7vsLPU
Video solution
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.